Karen flies her Cessna Skyhawk aircraft 360 miles with a tailwind of 10 miles per hour in the same time it could fly 300 miles with a headwind of 10 miles per hour. Find the speed of the aircraft.

Let suppose that aircraft travels at constant speed. From Mechanical Physics, we understand speed as travelled distance divided by time. In this case, the absolute speed of the aircraft is the sum of wind speed and speed of the aircraft relative to wind. That is:

Tailwind (Wind flows in the same direction of aircraft)

[tex]t = \frac{s_{T}}{v_{W,T}+v_{A/W,T}}[/tex] (Eq. 1)

Headwind (Wind flow in the direction opposite to aircraft)

[tex]t = \frac{s_{H}}{v_{W,H}+v_{A/W,H}}[/tex] (Eq. 2)

Where:

[tex]s_{T}[/tex], [tex]s_{H}[/tex] - Travelled distances on tailwind and headwind conditions, measured in meters.

[tex]v_{W,T}[/tex], [tex]v_{W,H}[/tex] - Tailwind and headwind speeds, measured in meters per second.

[tex]v_{A/W,T}[/tex], [tex]v_{A/W,H}[/tex] - Relative speeds of aircraft relative to tailwind and headwind, measured in meters per second.

We proceed to equalize both expressions to eliminate time:

[tex]\frac{s_{T}}{v_{W,T}+v_{A/W,T}} = \frac{s_{H}}{v_{W,H}+v_{A/W,H}}[/tex] (Eq. 3)

If we know that [tex]s_{T} = 360\,mi[/tex], [tex]s_{H} = 300\,mi[/tex], [tex]v_{W,T} = 10\,\frac{mi}{h}[/tex], [tex]v_{W,H} = -10\,\frac{mi}{h}[/tex] and [tex]v_{A/W,T} = v_{A/W,H}[/tex], then:

[tex]\frac{360}{v+10} = \frac{300}{v-10}[/tex]

The relative speed of the aircraft is now cleared:

[tex]360\cdot (v-10)=300\cdot (v+10)[/tex]

[tex]360\cdot v-3600 = 300\cdot v +3000[/tex]

[tex]60\cdot v = 7600[/tex]

[tex]v = 126.667\,\frac{mi}{h}[/tex]

The speed of the aircraft is 126.667 miles per hour.